A particle $A$ moves along the line, $y=30 \mathrm{~m}$ with a constant velocity, $\mathbf{v}$ parallel to $x$-axis. At the moment particle $A$ passes the $y$-axis, a particle $B$ starts from the origin with zero initial speed and a constant acceleration, $\mathbf{a}=0.40 \mathrm{~m} / \mathrm{sec}^2$. The angle between $\mathbf{a}$ and $y$-axis is $60^{\circ}$. If the particles $A$ and $B$ collide after sometimes, then the value of $|\mathbf{v}|$ will be

• A

  $2 \mathrm{~m} / \mathrm{s}$
• Correct answer
  B

  $3 \mathrm{~m} / \mathrm{s}$
• C

  $4 \mathrm{~m} / \mathrm{s}$
• D

  $5 \mathrm{~m} / \mathrm{s}$

Solution

As shown in the figure, two particles are moving with velocity $v$ and acceleration a.
Let the time of collision of two particles be $t$, then the equation of motion, $$ s=u t+\frac{1}{2} a t^2 $$ For particle $A$, $$ \begin{aligned} 30 \mathrm{~m} & =(0) t+\frac{a \cos 60^{\circ} t^2}{2} \\ \Rightarrow \quad 30 & =\frac{0.4}{2} \times \frac{1}{2} \times t^2 \\ t & =\sqrt{300} \end{aligned} $$ Hence, in time $t=\sqrt{300}$, both travelled equal distance in horizontal direction, so that the collision takes place. $$ \Rightarrow u t=\frac{1}{2} a \cos 30^{\circ} t^2\left(\because 30^{\circ}=90^{\circ}-60^{\circ}\right) $$ $$ \begin{aligned} & \Rightarrow \quad u=\frac{1}{2} \times 0.4 \times \frac{\sqrt{3}}{2} \times \sqrt{300} \\ & \Rightarrow \quad u=3 \mathrm{~m} / \mathrm{s} \\ & \end{aligned} $$ Hence, the correct option is (2).

Hence, the correct answer is option: 'B'.